IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

regi

the interior angle bisector of angle A for the triangle ABC whose co-ordinates of the vertices are

A(-8,5) ;B(-15,-19) and C(1,-7) has the equation ax + 2y + c =0.

find 'a' and 'c'

coolank2

The vertices of the triangle are A(-8,5), B(-15,-19) & C(1,-7).

Hence AB = 25, BC = 20 & CA = 15.

Now theres a prop of angle bisector that it divides the opposites side in the ratio of its arms. Hence AD divides BC in the ratio 5:3.

Hence, D(-5,-23/2).

Now, we can write the equation of AD in two point form.

It is (-13/3)x + 2y - (5/3) = 0

Hence a = (-13/3) and c = (-5/3)

vishnu_va

Find the equation of the angle-bisector,such that the coefficient of y is 2. Then compare the eq. with the given eq.

iitkgp_bipin

Too good coolank2.

Apply the property of the internal bisector :
If the internal bisector of A meets BC at D then AB/AC = BD/CD.
AB=25, AC=15
so BD/CD = 25/15 = 5/3

Now D can be found since it is known that D divides BC internally in the ratio 5:3.

Now you have two points A and D and can find the equation of line passing through them.

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